2 Composition. Invertible Mappings

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1 Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan Composition. Invertible Mappings In this section we discuss two procedures for creating new mappings from old ones, namely, the composition of mappings and invertible mappings. Composition of Two Mappings Composition is the combination of two or more mappings to form a single new mapping. Definition.1 Let α : S T and β : T U be two mappings. We define the composition of α followed by β, denoted by β α, to be the mapping for all x α. (β α)(x) = β(α(x)) Note carefully that in the notation β α the mapping on the right is applied first. See Figure.1 Figure.1 Example.1 Let α and β be given by the Venn diagram of Figure.. 1

3 while (α β)(x) = α(β(x)) = α(x 1) = (x 1) + = x x + 3 This example, shows that, in general, α β and β α need not be equal. In the following two theorems, we discuss the question of either composing two onto mappings or two one-to-one mappings. Theorem.1 Assume that α : S T and β : T U are two mappings. (a) If α and β are onto then the composition β α is also onto. (b) If β α is onto then β is onto. (a) The mapping β α is a mapping from S to U. So, let u U. Since β is onto then there is a t T such that β(t) = u. Now, since α is onto then there is an s S such that α(s) = t. Thus, given u U we can find an s S such that (β α)(s) = β(α(s)) = β(t) = u. This says that β α is onto. (b) Suppose now that β α is onto. Pick an arbitrary element u U. Since β α is onto then there is an s S such that β(α(s)) = u. Let t = α(s) T. Then β(t) = u. This shows that β is onto. Example.4 Consider the two mappings α : N N defined by α(n) = n and β : N N defined by { n+1 if n is odd β(n) = n if n is even Show that β and β α are onto but α is not. First, we show that β is onto. Let n N. If n is even then n is even and β(n) = n. If n is odd then n 1 is odd and β(n 1) = n. Thus, β is onto. One can easily check that β α = ι N. Since the identity map is onto then β α is onto. The mapping α is not onto since odd positive integers do not have preimages. 3

4 Theorem. Assume that α : S T and β : T U are two mappings. (a) If α and β are one-to-one then β α is also one-to-one. (b) If β α is one-to-one then α is one-to-one. (a) Suppose that α and β are one-to-one. Suppose that (β α)(s 1 ) = (β α)(s ) for some s 1, s S. This implies that β(α(s 1 )) = β(α(s )). Since β is one-to-one then α(s 1 ) = α(s ). Now since α is one-to-one then s 1 = s. Thus, β α is one-to-one. (b) Assume that β α is one-to-one. Suppose that α(s 1 ) = α(s ). Since β is a well-defined mapping then β(α(s 1 )) = β(α(s )). Since β α is one-toone then s 1 = s. This shows that α is one-to-one. Example.5 Consider the two mappings α : N N defined by α(n) = n and β : N N defined by { n+1 if n is odd β(n) = n if n is even Show that α and β α are one-to-one but β is not. Since α(n 1 ) = α(n ) implies n 1 = n and this in turns implies that n 1 = n then α is one-to-one. Since β α = ι N and ι N is one-to-one then β α is one-to-one. The mapping β is not one-to-one since β(1) = β() with 1. Example.6 Let α : S T, β : T U, and γ : U V be three mappings such that γ (β α) and (γ β) α are well defined. Show that γ (β α) = (γ β) α Note first that the mappings γ (β α) and (γ β) α have the same codomain V. The following argument shows that γ (β α) and (γ β) α have the 4

5 same range. Let s be in S then [γ (β α)](s) = γ((β α)(s)) = γ(β(α(s))) = (γ β)(α(s)) = [(γ β) α](s) Invertible Mappings In this section we consider special kind of mappings which have the property that for each output value we can work our way backwards to find the unique input that produced it. Let α : S T and β : T S be two given mappings. Definition. We say that β is an inverse of α if and only if β α = ι S and α β = ι T. In this case we say that β is invertible. An invertible mapping has a unique inverse as shown in the next theorem. Theorem.3 If α : S T is invertible then its inverse is unique. Suppose that α 1 : T S and α : T S are two inverses of α. Then from Definition. we have α 1 α = α α = ι S and α α 1 = α α = ι T. We want to show that the mappings α 1 and α are equal. That is, we must show that α 1 (t) = α (t) for each t T. Indeed, Thus, α 1 = α. α 1 (t) = ι S (α 1 (t)) = (α α)((α 1 (t)) = α ((α α 1 )(t)) = α (ι T (t)) = α (t) Definition.3 We denote the unique inverse of a mapping α by α 1. 5

6 Example.7 Show that the mapping α in Example.1 is invertible and find its inverse. Solution The inverse of α is defined by α 1 (1) = y, α 1 () = x, α 1 (3) = z. One can easily check that α α 1 = ι T and α 1 α = ι S where S = {x, y, z} and T = {1,, 3}. Looking closely at the Venn diagram we see that α 1 is gotten by reversing the direction of the arrows under α. The following theorem characterizes those mappings that are invertible. Theorem.4 A mapping α : S T is invertible if and only if α is one-to-one and onto. Suppose first that α is invertible with inverse α 1 : T S. We will show that α is both one-to-one and onto. To see that α is one-to-one, we assume that α(s 1 ) = α(s ), where s 1, s S, and show that s 1 = s. Indeed, s 1 = ι S (s 1 ) = (α 1 α)(s 1 ) = α 1 (α(s 1 )) = α 1 (α(s )) = (α 1 α)(s ) = ι S (s ) = s Next, to show that α is onto we pick an arbitrary member t in T and show that there is an s in S such that α(s) = t. Indeed, since t is in T then t = ι T (t) = (α α 1 )(t) = α(α 1 (t)) = α(s) where s = α 1 (t) S. This shows that α is onto. Conversely, suppose that α is one-to-one and onto. We will find a mapping β : T S such that α β = ι T and β α = ι S. Let t T. Since α is onto then there is an element s S such that α(s) = t. s is unique, for if s S is such that α(s ) = t then α(s) = α(s ). But α is one-to-one so that s = s. Hence, for each t T there is a unique s S such that α(s) = t. Define β : T S by β(t) = s. Then β satisfies the following properties: and (α β)(t) = α(β(t)) = α(s) = t, t T (β α)(s) = β(α(s)) = β(t) = s, s S. That is, α β = ι T and β α = ι S. According to Definition., α is invertible 6

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